Disclaimer: I posted recently to MSE a similar question that still remains unanswered. This question is not a duplicate and instead is about another approach suggested by a comment on the original post to the proof of the question .
We assume that $f$ is an entire function of finite order and that $|f|\le 1$ on the boundary of the strip $S= \{z\in \mathbb{C}: \operatorname{Re} z\ge 0, -1< \operatorname{Im} z<1\}$. We let $z=x+iy$. I would like to use a modified version of the maximum modulus principle to show that $|f(z)|\le 1$ on all of $S$.
I begin by noting that since $f$ is of finite order, there exists $C, m>0$ such that $|f(z)|\le C e^{|z|^m}$ for all $z\in\mathbb{C}$. As is outlined in wikipedia page of the Phragmén-Lindelöf principle, I now consider the still entire function $h_\varepsilon(z)=f(z)e^{-\varepsilon \exp(z)}$. We have that $$\left|h_\varepsilon(z)\right|\le |f(z)|e^{-\varepsilon \cos(y)\exp(x)}\le |f(z)|e^{-\varepsilon C\exp(x)},$$ where $C=\cos(1)>0$. Because of the fast decay of $e^{-\varepsilon C\exp(x)}$, we can pick large enough $R>0$ so that (1) $|h_\varepsilon|\le 1$ on the boundary of $S\cap D(0,R)$ with $D(0,R)$ the disk of radius $R$, and (2), $|h_\varepsilon|\le 1$ on $S\cap D(0,R)^c$ (I denote the complement of $X$ as $X^c$). By the maximum principle applied to (1), we have that $|h_\varepsilon|\le 1$ on all of $S\cap D(0,R)$ and (2) says the same estimate holds on the rest of $S$. Thus, $|h_\varepsilon|\le 1$ holds on all all of $S$ for every $\varepsilon>0$ and we may now let $\varepsilon \to 0$ to conclude that $|f(z)|\le 1$ on $S$ as desired.
Is the proof correct and may it be improved anywhere?